Skip to main content

Full text of "Treatise On Analysis Vol-Ii"

See other formats


17.    Let A be a Banach algebra with unit element <?, and let x>->jt* be an involution on A
which is not necessarily continuous.

(a)    Show that the radical SR of A (Section 1 5.2, Problem 7) is a self-adjoint subset of A.
Passing to the quotient, it follows that the involution on A defines an involution
(also written z*~*z*) on A/8ft.

(b)    Let a = a* be a self-adjoint element of A, such that SpA(#) is contained in the
open half-plane & > 0. Then there exists an element b e A, which is the limit of a
sequence (un) of self-adjoint elements which are polynomials in a, and which satisfies
b2 = a. (Write a  oc(e  (e  a"^)) for a suitable a > 0, and use Problem ll(f)
and Section 15.2, Problem 15.) Show that b and b* commute, and that we may there-
fore write b = u -f iv, where u and v are self-adjoint and commute with each other and
with a. Deduce that uv = 0 and u2 ~ v2 = a.

(c)    Let C be a commutative Banach subalgebra of A which contains e, a, b, b* and
is such that Spc(&) = SpA(6) (Section 15.2, Problem 15). Let C' be a commutative
Banach subalgebra of A containing the image of C under the homomorphism 9 : x*-*x*
of C into A. If 9l7 is the radical of C' and vr : C' -> C'/ST the canonical homomorphism,
then the homomorphism TT  99 : C-^C'/B' is continuous (Section 15.3, Problem 20).

Deduce that v =  (b*  b) e SR', and thence that SpA(<a) = SpA(w2). (Observe that

SpA(<2) = SpC'/$R/(7r(w2  v2)) = Spc'/W'W"2)).) Consequently u is invertible, v =
w~1wt? = 0, and finally & = w is self-adjoint. Furthermore, if Sp(a) is contained in
]0, 4- oo [, then the same is true of Sp(6).

18.    With the hypotheses of Problem 17, put pA(x) = (p(je*x))1/2 (also denoted by p(x)).
We have X**) = P(X) ; also if s$ is the radical of A and if TT : A -> A/sJt is the canonical
homomorphism, then ^A/wC^W) = PA(X).

The involution jci-^x* on A is said to be hermitian, and the involutive algebra A
is said to be hermit Ian, if Sp(a) c R for all self-adjoint elements a E A. If A is com-
mutative, an equivalent condition is that every character of A is hermitian (Problem 2).

Assume for the rest of this Problem that A is hermitian.

(a) Let x e A be such that p(x) < I . Show that (e -f x*)(e  x) is invertible. (Remark
that e  x*x  w2, where w is self-adjoint and invertible, by virtue of Problem 17;
hence show that the spectrum of w~l(x*  x)w"x is contained in z'R.) Deduce that

for all x e A (Ptdk's inequality),

Consequently, if x is normal, we have p(x) p(x). (Remark that if x, y are two
elements of a Banach algebra which commute, then p(xy) ^p(x)p(y), by (15.3.4).)
In particular, if x is unitary, then p(x)  I and Sp(x) c U.

(b)    Deduce from (a) that for each pair a, b of self-adjoint elements of A we have
p(ab) <; p(a)p(b). (Observe that p(ab) <>p(ab)  (p(a2b2))lf2 <; ||a2||1/2  I|62II1/2.) For
all x, y e A we have p(xy) < p(x)p(y), and the radical of A is the set of all x e A such
that p(x) = 0.

(c)    Suppose that a, b are self-adjoint elements of A such that Sp(<a) and Sp(6) are
contained in [0, -f oo [. Show that Sp(a -f- b) is contained in [0, + oo [. (Write

with p(u) < I and p(v) < 1 .) Deduce that for any two self-adjoint elements cz, b of A,
we have p(a + b) < p(a) + p(b). Show also that Sp(;t*je + xx*) <= [0, + co [ for all
x e A. the real vector subspace H of A con-